Practicing Success
The equation $2^{|x^2-12|}=\sqrt{e^{|x|\log 4}}$ has |
no real solution only two real solutions whose sum is zero only two real solutions whose sum is not zero four real solutions whose sum is zero. |
four real solutions whose sum is zero. |
We have, $2^{|x^2-12|}=\sqrt{e^{|x|\log 4}}$ $⇒2^{|x^2-12|}=\sqrt{e^{\log 4^{|x|}}}$ $⇒2^{|x^2-12|}=\sqrt{4^{|x|}}$ $⇒2^{|x^2-12|}=2^{|x|}⇒|x^2-12|=|x|$ Now, $x^2-12=0⇒ x=±2\sqrt{3}$. So, we have the following cases. CASE I When $x ≥2\sqrt{3}$ In this case, we have $|x^2-12|=x^2-12$ and $|x|= x$ $∴|x^2-12|=|x|$ $⇒x^2-12=x$ $⇒x^2-x-12=0$ $⇒(x-4) (x+3)=0⇒ x=4$ $[∵x≥2\sqrt{3}]$ CASE II When $x ≤-2\sqrt{3}$ In this case, we have $|x^2-12|=x^2-12$ and $|x|= x$ $∴|x^2-12|=|x|$ $⇒x^2-12=-x$ $⇒x^2+x-12=0$ $⇒(x+4) (x-3)=0⇒ x=-4$ $[∵x≥-2\sqrt{3}]$ CASE III When $-2\sqrt{3}<x<0$ In this case, we have $|x^2-12|=-(x^2-12)$ and $|x|= x$ $∴|x^2-12|=|x|$ $⇒-(x^2-12)=-x$ $⇒x^2-x-12=0$ $⇒(x-4) (x+3)=0⇒ x=-3$ $[∵-2\sqrt{3}<x<0]$ CASE IV When $0≤x<2\sqrt{3}$ In this case, we have $|x^2-12|=-(x^2-12)$ and $|x|= x$ $∴|x^2-12|=|x|$ $⇒-(x^2-12)=x$ $⇒x^2+x-12=0$ $⇒(x+4) (x-3)=0⇒ x=3$ $[∵0≤x<2\sqrt{3}]$ Hence, the four real solutions are ± 4, ± 3. Clearly, their sum is zero. |